Just a quick note, showing a polynomial of degree 2 or 3 has no roots in a particular field is enoughto say it is irreducible over that field. In particular you ruled them all out over the rationals with rational root test, so you are done. No need to worry about another possible factorization.

Last edited by Gamma; April 19th 2009 at 11:30 AM.
Reason: Didn't show up where I thought it would

Right

Oh, don't apologize, to be honest I didn't even notice the degrees didn't add up.

I hope that you mean your contradiction will be that e cannot be in as you have shown there are not roots, and therefore cannot have a linear term, thus is necessarily irreducible if it has no linear term factors because it has degree 3. I just don't want you to multiply anything out and start solving systems of equations or anything, lol. Just want to make sure we are on the same page.

Last edited by Gamma; April 19th 2009 at 01:12 PM.
Reason: clarification